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I want to show that

$ Y(t)=\cos(At)$ and $Y(t)=\sin(At)$ satisfy the equation

$$Y''+A^2Y=0 $$

subjected to the initial conditions $Y(0)=I, Y'(0)=0$ and $Y(0)=0, Y'(0)=I$ respictively

where $A$ is an constant $n \times n$ matrix..

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Something is strange here...if $Y(t)=\cos(At)$, then $Y(0)$ is never going to be the identity matrix. It will be a matrix full of ones. That is, assuming $t$ is a scalar...? –  icurays1 Jan 14 '13 at 23:48
4  
@icurays1 It's not element-wise cosine. See the Taylor expansion below in Mhenni's answer. –  John Moeller Jan 14 '13 at 23:52
    
Well. That's embarrassing. –  icurays1 Jan 14 '13 at 23:53

3 Answers 3

up vote 1 down vote accepted
+50

If $$X(t):=\cos(A t):=\sum_{k=0}^\infty{\cos(k\pi/2)\over k!} A^k t^k$$ then $X(0)=I$ and $$X'(t)=\sum_{k=1}^\infty{\cos(k\pi/2)\over k!} k A^k t^{k-1}=-A\ \sum_{k=0}^\infty{\sin(k\pi/2)\over k!} A^k t^k=-A\ \sin(At)\ ;$$ in particular $X'(0)=0$.

Furthermore, if $$Y(t):=\sin(A t):=\sum_{k=1}^\infty{\sin(k\pi/2)\over k!} A^k t^k$$ then $Y(0)=0$ and $$Y'(t)=\sum_{k=1}^\infty{\sin(k\pi/2)\over k!} k A^k t^{k-1}=A\ \sum_{k=0}^\infty{\cos(k\pi/2)\over k!} A^k t^k=A\ \cos(At)\ ;$$ in particular $Y'(0)=A$.

It follows that $X''(t)=-A Y'(t)=-A^2 X(t)$; so $X(\cdot)$ satisfies the given differential equation as well as the initial conditions $X(0)=I$, $X'(0)=0$.

For $Y(\cdot)$ it's different: This function satisfies the differential equation as well as the initial condition $Y(0)=0$. But if we want $Y'(0)=I$ instead of $Y'(0)=A$ we have to replace $Y(\cdot)$ by the new solution $$\hat Y(t):=A^{-1}Y(t)=A^{-1}\sin(A t)\ .$$ When $A$ happens to be singular we seem to be in trouble. But there is remedy: Since the sine series begins with the linear term we can write $\hat Y(t)$ without explicit appearance of $A^{-1}$: $$\hat Y(t)=\sum_{k=1}^\infty{\sin(k\pi/2)\over k!} A^{k-1} t^k= t\ {\rm sinc}(A t)\ .$$

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Hints:

1) Use the power series representation

$$ Y(t)=\sin(At) = \sum_{k=0}^{\infty}(-1)^kA^{2k+1}\frac{t^{2k+1}}{(2k+1)!}, $$

and same for $\cos(At)$.

2) If $Y_1$ and $Y_2$ are solutions of the ode, then

$$Y := c_1Y_1+c_2 Y_2$$

is a solution too.

Added

$$ Y(t)=\sin(At) = \sum_{k=0}^{\infty}(-1)^kA^{2k+1}\frac{t^{2k+1}}{(2k+1)!} $$

$$\implies Y'(t)=\sum_{k=0}^{\infty}(-1)^kA^{2k+1}(2k+1)\frac{t^{2k}}{(2k+1)!} $$

$$\implies Y''(t)=\sum_{k=1}^{\infty}(-1)^kA^{2k+1}(2k+1)(2k)\frac{t^{2k-1}}{(2k+1)!}=\sum_{k=1}^{\infty}(-1)^kA^{2k+1}\frac{t^{2k-1}}{(2k-1)!}. $$

Shifting the index in the above sum gives

$$ \sum_{k=0}^{\infty}(-1)^{k+1}A^{2(k+1)+1}\frac{t^{2(k+1)-1}}{(2(k+1)-1)!}= -A^2 \sum_{k=0}^{\infty}(-1)^{k}A^{2k+1}\frac{t^{2k+1}}{(2k+1)!}$$

$$ \implies Y''(t) = -A^2 \sin(At). $$

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How can you say $$Y'(t)=\sum_{k=0}^{\infty}(-1)^kA^{2k+1}(2k+1)\frac{t^{2k}}{(2k+1)!} \implies Y''(t)=\sum_{k=1}^{\infty}(-1)^kA^{2k+1}(2k+1)(2k)\frac{t^{2k-1}}{(2k+1)!} $$ How does $k=0$ change into $k=1$ ? –  MSKfdaswplwq Jan 15 '13 at 17:38
    
Hmm... I see, if k=0, then the whole term vanishes. –  MSKfdaswplwq Jan 15 '13 at 18:22
    
What do I have to do with the initial conditions? I can show that $Y''(t)=0A^2sin(At)$... –  MSKfdaswplwq Jan 15 '13 at 20:57
    
You need to find the constants $c_1$ and $c_2$. –  Mhenni Benghorbal Jan 15 '13 at 22:46
    
How can you do that? –  MSKfdaswplwq Jan 16 '13 at 12:57

I found the shifting of indices to be highly confusing so I propose a method that does not rely on that.

We will first show that $Y(t)=\cos (At)$ solves the initial value problem $Y''+A^2 Y=0$ where $Y(0)=I$ and $Y'(0)=0$. Because $\cos At$ is defined as the infinite sum $\sum\limits_{k=0}^{\infty} (-1)^{k}\frac{A^{2k}t^{2k}}{(2k)!}$ and differentiation is linear we can differentiate term by term to get:

\begin{eqnarray*} Y''(t)&=&\frac{d^2}{dt^2}\left[ I -\frac{A^2t^2}{2!}+\frac{A^4t^4}{4!}-\frac{A^6t^6}{6!}+\ldots \right]\\ &=&\left[ -2\frac{A^2}{2!} +4\cdot 3 \frac{A^4 t^2}{4!} -6\cdot 5 \frac{A^6 t^4}{6!} +\ldots \right]\\ &=&-A^2\left[ I -\frac{A^2t^2}{2!}+\frac{A^4t^4}{4!}-\frac{A^6t^6}{6!}+\ldots \right]\\ &=& -A^2 \sum\limits_{k=0}^{\infty} (-1)^k \frac{A^{2k}t^{2k} }{(2k)!}=-A^2Y(t) \end{eqnarray*}

So $Y(t)$ satisfies $ Y''+A^2Y=0$. Also $Y(0)=\cos(0)=\sum\limits_{k=0}^{\infty} (-1)^{k}\frac{0^{2k}}{(2k)!}=I$ (The $0^{2k}$ in the summation is the zero matrix $[A\cdot t]_{t=0}=0$ and $0^0=I$, all higher powers of $0$ are of course $0$). And for $Y'(0)$ we find

\begin{eqnarray*} Y'(0)&=&\frac{d}{dt}\left[ I -\frac{A^2t^2}{2!}+\frac{A^4t^4}{4!}-\frac{A^6t^6}{6!}+\ldots \right] _{t=0}\\ &=& \left[ -2\frac{A^20}{2!}+4\frac{A^40^3}{4!}-6\frac{A^60^5}{6!}+\ldots \right] =0 \end{eqnarray*} (here the $0^k$ is just the scalar $t=0$ )

For $Y(t)=\sin(At)$ with $\sin(At):=\sum\limits_{k=0}^{\infty} (-1)^{k}\frac{A^{2k+1}t^{2k+1}}{(2k+1)!}$ we have:

\begin{eqnarray*} Y''(t)&=&\frac{d^2}{dt^2}\left[ At -\frac{A^3t^3}{3!}+\frac{A^5t^5}{5!}-\frac{A^7t^7}{7!}+\ldots \right]\\ &=&\left[ -3\cdot 2\frac{A^3t}{3!} +5\cdot 4 \frac{A^5 t^3}{5!} -5\cdot 6 \frac{A^7 t^5}{7!} +\ldots \right]\\ &=&-A^2\left[ At -\frac{A^3t^3}{3!}+\frac{A^5t^5}{5!}-\frac{A^7t^7}{7!}+\ldots \right]\\ &=& -A^2 \sum\limits_{k=0}^{\infty} (-1)^{k}\frac{A^{2k+1}t^{2k+1}}{(2k+1)!}=-A^2Y(t) \end{eqnarray*}

So $Y(t)=\sin(At)$ also satisfies $Y''(t)+A^2Y(t)=0$. Here however $Y(0)=\sin( 0)=\sum\limits_{k=0}^{\infty} (-1)^{k}\frac{0^{2k+1}}{(2k+1)!}=0$ (again using $0$ as the zero matrix in the summation, and now the lowest power of $0$ is $0^1$ so here we have an infinite sum of zeros). And

\begin{eqnarray} Y'(0)&=&\frac{d}{dt}\left[ At -\frac{A^3t^3}{3!}+\frac{A^5t^5}{5!}-\frac{A^7t^7}{7!}+\ldots \right] _{t=0}\\ &=& \left[ A -3\frac{A^30^2}{3!}+5\frac{A^50^4}{5!}-7\frac{A^70^6}{7!}+\ldots \right] =A \end{eqnarray} I am guessing this is what you meant because even if we consider a scalar $A$ then $\left[\frac{d}{dt}\sin(At)\right]_{t=0}=A\cos(0)=A$.

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